Question

Let $A = \\{ 0,1,2 \\}$, $B =\\{ 4,2,0 \\}$ and $f,g$ : $A \rightarrow B$ be the functions defined by $f(x) = x^2-x$ and $g(x) = 2|x-\frac{1}{2}|-1$ Then,

• A. $f = g$
• B. $f = 2g$
• C. $g = 2f$
• D. None of these

Answer 1

Answer: (A)

$f = g$

Answer 2

We have A=\left\\{1, 0, 1, 2\right\\}
B=\left\\{4, 2, 0, 2\right\\}
and $f, g : A \rightarrow B$
Now, $f \left(x\right)=x^{2}-x$
$\Rightarrow f \left(0\right)=0-0=0$
and $f \left(1\right)=1-1=0\ldots\left(i\right)$
and $f \left(2\right)=2^{2}-2=2$
Also, $g\left(x\right)=2\left|x-\frac{1}{2}\right|-1$
$\Rightarrow g \left(0\right)=2\left|\frac{-1}{2}\right|-1=1-1=0$
and $g\left(1\right)=2\left|1-\frac{1}{2}\right|-1$
$=1-=0 \ldots\left(ii\right)$
and $g \left(2\right)=2\left|2-\frac{1}{2}\right|-1$
$=\frac{2.3}{2}-1=2$
Hence from Eqs. $\left(i\right)$ and $\left(ii\right)$, we can say that
$f\left(x\right)=g\left(x\right)$
$\Rightarrow f=g$